Lecture 16 Air Pollution (3)

CVEN 301 Introduction to Environmental EngineeringFall 2007

Dr. Qi YingDepartment of Civil Engineering

Atmospheric Dispersion Modeling (2)

Plume Rise

A plume of hot gas emitted vertically rises due to its Momentum Buoyancy

Plume Rise

V0

V

Loses momentum due to entrainment

Ts>>T

Δh

H

Ts”~Ta

Ts’>T

Gradually loses buoyancy and bends over

Parameters affect plume rise Plume rise depends on both plume

and ambient parameters Plume and stack parameters Exit velocity Stack diameter Gas temperature Gas molecular weight

Ambient air parameters Stability Wind speed Temperature

Holland’s Simple Equation Includes stack and plume parameters Does not take atmospheric stability into

consideration

21.5 2.68 10s s s aa s

s

v d T Th p d

u T

vs = stack exit velocity (m/s)ds = stack diameter (m)u = wind velocity (m/s)pa = atmospheric pressureTs = stack temperature (K)Ta = ambient temperature

Holland’s Simple Equation

For large power plants, the heat emission rate (QH) is usually reported instead of stack temperature

1.5 9.6s s Hv d Qh

u u

vs = stack exit velocity (m/s)ds = stack diameter (m)u = wind velocity (m/s)QH = heat emission rate (MW)

Briggs Plume Rise Equations

It is the current EPA recommend method for plume rise calculation

It has better performance for thermally dominated plume (buoyancy >> momentum)

Plume rise can be estimated as a function of downwind distance

http://www.air-dispersion.com/briggs.html

2

4s a

s ss

T TF gv d

T

Buoyancy factor (F):

Stability parameter (s):

0.02 for stability class E

0.035 for stability class F

a

a

gS

T

gS

T

Δh

h

xf

Example – Plume Rise

For Class D stability, calculate the final plume rise using Briggs equations from a power plant stack, given the following information

vs = 20 m/sds = 5 mU = 6 m/sTs = 400 KTa = 280 K

Example – Plume Rise

Calculate Buoyancy Factor

F>55, calculate downwind distance where maximum plume rise happens

Final plume rise

2 2 400 2809.81 20 5 525.5

4 4 280s a

s ss

T TF gv d

T

0.4 0.4119 119 525.5 1458fx F m

1/3 2/31/3 2/3 1 1.6 525.5 1458

1.6 2776fh F x U m

Wind Speed as a Function of Height

The wind speed (u2) at stack height (z2) can be estimated using surface wind measurement(u1 @ z1):

22 1

1

pz

u uz

Stability urban ruralA 0.15 0.07B 0.15 0.07C 0.2 0.1D 0.25 0.15E 0.3 0.35F 0.3 0.35

Dependence of p as a function of stability and surface roughness

Wind speed example

Calculation wind speed at 477m if the wind speed at 10m above surface is 2 m/s. Assume neutral condition in urban area.

U477=U10*(477/10)0.25

=2*2.62=5.3 m/s

Maximum Ground Surface Concentration

The surface concentration can be derived by setting z=0 in the equation:

The maximum ground concentration must occur at y=0

2 2

2 2( , ,0) exp exp

2 2y z y z

E y HC x y

S S U S S

(H=h+Δh)

2

2( ,0,0) exp

2y z z

E HC x

S S U S

Maximum Ground Concentration (Neutral)

Stack

2 2

2 2( , ,0) exp exp

2 2y z y z

E y HC x y

S S U S S

H=25mStability class = DE=1g/sU=1m/s

Maximum Ground Concentration (Unstable)

2 2

2 2( , ,0) exp exp

2 2y z y z

E y HC x y

S S U S S

H=25mStability class = AE=1g/sU=1m/s

Stack

Summarize – Gaussian Dispersion Problem

Determine stability class Calculate plume rise Calculate wind speed Calculate Sy, Sz Calculate pollutant concentration

Example Determine the pollutant surface concentration

at 2 meter above surface, 400 meters directly downwind of the stack. Assume stability class D, wind speed 2m/s at effective stack height, pollutant emission rate 1g/s and an effective stack height of 20 m. Also assume that the pollutant is perfectly reflected when it hits the ground.

Example

Solution:u=2m/s, H=20m, Stability Class=DPosition to calculation concentration

(400,0,2) 2 22

2 2 2( , , ) exp exp exp

2 2 2 2y z y z z

z H z HE yC x y z

S S U S S S

2 22

2 2 2

2 2

2 2

2 20(400,0,2) exp exp exp

2 2 2 2

2 2 exp exp

2 2 2

y z y z z

y z z z

H HEC

S S U S S S

H HE

S S U S S

Example

Calculate Sy, Sz:

Sy = a*x0.894 Sz = c*xd + fx<1km x>1km

Stability a c d f c d fD 68 33.2 0.725 -1.7 44.5 0.516 -13E 50.5 22.8 0.678 -1.3 55.4 0.305 -34F 34 14.35 0.74 -0.35 62.6 0.18 -48.6

0.894

0.678

400 1

68 (400 /1000) 30

22.8 (400 /1000) 1.3 11

y

z

x m km

S m

S m

Example

Calculate concentration

2 2

2 2

2 2

2 2

6 3 3

2 2(400,0,2) exp exp

2 2 2

2 20 2 201 exp exp

2 30 11 2 2 11 2 11

8.56 10 / 8.56 /

y z z z

H HEC

S S U S S

g m g m

Puff Release Sometime we need to

determine pollutant concentrations downwind due to an instantaneous release

The plume is advected downwind as a “puff”

Puff concept

x

y

Mass=mt1=U/x1

Mass=mt2=U/x2

Mass=mt3=U/x3

-Pollutant concentration decreases due to dispersion in all directions.-The total mass in the puff remains unchanged.

Puff concentration

The concentration of pollutant (C) at ground surface (x,y) at any given time (t) can be calculated by

22 2

1.5

1( , ,0, ) exp

22( ) x y zx y z

m x Ut y HC x y t

S S SS S S

Sx, Sy Sz = dispersion parameters (m)U = wind speed at plume release point (m/s)t = time after plume release (s)m = amount of pollutant released (kg)H = height where the puff is released (m)

Dose

The amount of pollutant received during pollutant exposure (grams.second/m3)

At ground level

0( , , ) ( , , , )D x y z C x y z t dt

2 21

( , ,0) exp2y z y z

m y HD x y

S S U S S

Coefficient Stability a b Unstable 0.14 0.92

Sy Neutral 0.06 0.92 Stable 0.02 0.89 Unstable 0.53 0.73

Sz Neutral 0.15 0.7 Stable 0.05 0.61

b by zS ax S ax

Sy and Sz are different from plume dispersion equations

(x in meters)

Ground Level Dose

Neutral Condition M=1kgU=1m/sH=2m

Highway Air Pollution

Emissions from highway account for majority of the CO, NOx and VOC in urban areas

Inappropriate arrangement of highways lead to local “hotspot” in air quality

Gaussian dispersion model can be applied to highway segments to predict pollutant concentrations downwind

Finite Length Line Source (FLLS)

A highway section with uniform emission rate can be modeled as a finite line source

U

Wind directionx axis

y axis

Plume from a differential length y2

y1

y

Sy=f(x)

dy

Receptor at(x,y0)

y0

y0-y

Calculate FLLS concentration

Steady State q = Emission factor (kg/s.m)

Emission rate from a differential length = q.dy (kg/s)

Concentration at receptor due tothe differential emission

Integrate over the entire length:

2 20

0 2 2( , ,0) exp exp

2 2y z y z

y yqdy HdC x y

US S S S

Ux axis

y axisy2

y1

ydy

Receptor at(x,y0)

y0y0-y

2

10 0( , ,0) ( , ,0)

y

yC x y dC x y